That means the graph passes point (0,0).
f'(x)=0 for x=-2
That means the graph is horizontal at x = -2.
f"(x)=0 for x=1
That means the concavity of the graph changes at x = 1. There is an inflection point at x=1. The graph there needs not be horizontal, though.
f'(x)>0 for x<-2
The slope of the graph is positive to the left of x = -2.
So if at x = -2, the graph is horizontal, then there is maximum point at x = -2.
f'(x)<0 for -2<x<2
From x = -2 up to x = 2, the slope is negative.
f'(x)=0 for x>2
The graph is horizontal to the right of x = 2. Horizontal up to before infinity.
f"(x)<0 for x<1
The concavity of the graph is downwards to the left of x = 1.
That means from negative infinity up to x=1, the graph is like a vertical parabola that opens downward, whose vertex is at x = -2.
f" (x)>0 for 1<x<2
The concavity of the graph is upwards between x=1 and x=2. Yet, the slope there is still negative.
Okay, if you are still with me, the graph is like a weak or tipsy number two.
From after negative infinity up to x=1 it looks like a parabola with its highest point at x = -2.
It passes through the origin (0,0), so the highest point of the "parabola" is above the x-axis.
Then at x=1 it changes concavity--it faces upward, but it is still going down.
Then at x=2 it becomes horizontal until it reaches very close to positive infinity.